Date: 10/01/2002 at 13:21:47
From: Sarah
Subject: Mathematics
Just tell me the equation for that - any possible EQUATION.

Date: 10/01/2002 at 14:42:43
From: Doctor Ian
Subject: Re: Mathematics
Hi Sarah,
The reason you're given problems to solve isn't so you can just find
the answers and give them back to the teacher. The teacher already
_knows_ the answers.
The reason you're given problems to solve is so you can learn how to
approach problems that you've never seen before, and don't know how to
solve. I'm trying to help you learn to do that.
You're supposed to find two consecutive odd numbers whose product is
255. One way to do that is to find _all_ the factors of 255, and see
which ones are consecutive odd integers.
Now, 255 is divisible by 5:
255 = 5 * 51
and 51 is divisible by 3:
255 = 5 * 3 * 17
and all of these are prime factors. So there are only a few ways to
multiply numbers together to get 255.
1 * (3 * 5 * 17)
3 * (5 * 17)
5 * (3 * 17)
17 * (3 * 5)
Does one of these fit the problem description?
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/

Date: 10/02/2002 at 10:02:24
From: Sarah
Subject: Mathematics
In an examination I simply don't have the time to do factorisation!

Date: 10/02/2002 at 10:56:43
From: Doctor Ian
Subject: Re: Mathematics
Hi Sarah,
Actually, the way I showed you is _faster_ than setting up and solving
an equation. Here's how that would look:
The product of two consecutive odd numbers is 255.
What are the numbers?
If the first number is n, the second is n+2. Their
product is
n(n+2) = 255
n^2 + 2n - 255 = 0
(n + a)(n - b) = 0
which means that
a - b = 2
a * b = 255
So now you have two equations to solve, instead of one.
And you have to solve the second equation by finding the
factors of 255, so you can see which ones differ by 2.
(If you look carefully, you'll see that you're really right
back where you started.)
All I'm saying is, since you're going to have to find the factors of
255 _anyway_, why not just _start_ there, and then pick the factors
that fit the problem description?
It's often the case for a problem on a test that there are at least
two ways to do the problem. The first way is to set up an equation and
solve it, which works (if you set up the right equation), but which
takes a lot of time. The second way is to try to look at the problem
from the point of view of the person who made it up.
If you wanted to _create_ a problem like this, you wouldn't do it by
setting up the equation. You'd do it by picking a pair of consecutive
odd numbers, like 7 and 9, or 11 and 13, or 15 and 17. So all you have
to do to _solve_ the problem is find those numbers. Prime factoring is
a quick way to do that, which is why your teachers have tried to get
you to learn it over the years.
One of the quickest ways to factor a number is to make use of
divisibility rules. The first one I used was very simple: Every number
that is divisible by 5 must end in either 0 or 5:
5, 10, 15, 20, 25, 30, 35, ...
So I know that 255 must be divisible by 5, and I factor a 5 out of it:
51
_____
5 ) 255
Note: This isn't actually the way I do the division; I do it this way:
5 * 40 is 200,
so the answer is 40, plus however many times 5 goes into 55,
which is 11,
so the answer is 40 + 11 = 51.
In other words, I use the distributive property,
(1/5)*255 = (1/5)*200 + (1/5)*55
to do the division. This way, I don't need to use paper to do it.
So now I have
255 = 5 * 51
Another very common divisibility rule is that if the digits of a
number add up to something divisible by 3, then the number is
divisible by 3:
Number Sum of digits
------ -------------
3 3
6 6
9 9
12 1 + 2 = 3
15 1 + 5 = 6
18 1 + 8 = 9
21 2 + 1 = 3
:
2196 2 + 1 + 9 + 6 = 18 => 1 + 8 = 9
The reasons for this are a little complicated, but the trick itself is
pretty easy to apply. Anyway, I can use this trick to see that 51 is
divisible by 3, so I can factor a 3 out of it:
17
____
3 ) 51
(As before, I don't actually use long division; I pick something I
know is close, e.g., 3 * 15 = 45, so then I just have to divide 3 into
6, and add the two quotients: 15 + 2 = 17.)
So now I have
255 = 3 * 5 * 17
And I know that 17 is prime. So now I know that _any_ factor of 255
must be the product of the prime factors, which means I just have to
try to construct two consecutive odd numbers from these. One of the
numbers will clearly be 17, which leaves 3*5, or 15, as the other
number.
A third very useful rule is that any number that is divisible by 2
must end in 0, 2, 4, 6, or 8. Our FAQ has a whole section on
divisibility rules. A lot of them are pretty specialized, but the
rules for divisibility by 2, 3, 5, and 9 are worth learning by heart.
I know it must seem at times that math class is just about turning you
into a kind of robot that sets up and solves equations. But that's not
how it's supposed to be! Math is a set of tools that are designed to
let you get right to the heart of a problem, and solve it with a
minimum of effort.
And so there are two parts to learning math: the first is learning to
use the tools; the second is learning to recognize which tools to use
in a given situation. Unfortunately, in many math classes, almost all
the effort is directed at the first part, when in fact, it's the
second part that's most important.
In this case, it's as if you're trying to open a door that's been
locked. You can get a big hammer and try to knock down the door (e.g.,
set up and solve an equation); or you can check above the door frame
to see if someone left the key there (e.g., check the prime factors to
see if there is an obvious answer).
And sometimes you can just look for an open window, bypassing the door
completely. In this case, you could do something like this:
10^2 is 100, so 9 * 11 would be too small.
20^2 is 400, so 19 * 21 would be too big.
255 ends in 5, so one of the numbers has to be 15.
So it has to be 13 * 15, or 15 * 17.
When you approach it this way, solving a math problem is less like
what a robot does, and more like what a detective does, and it's when
you start to see it that way that it becomes fun, instead of boring.
Anyway, I hope this helps. Write back if you'd like to talk more
about this, or anything else.
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/